Ex.7.3 Q2 Coordinate Geometry Solution - NCERT Maths Class 10

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Question

In each of the following find the value of ‘\(k\)’, for which the points are collinear.

  • \((7, -2), (5, 1), (3, -k)\)
  • \((8, 1), (k, -4), (2, -5)\)

Text Solution

Reasoning:

Three or more points are said to be collinear if they lie on a single straight line .

What is Known:

The \(x\) and \(y\) co-ordinates of the points.

What is  Unknown:

The value of ‘\(k\)’, for which the points are collinear.

Solution:

(i) Given,

  • Let A \((x_1, y_1) = (7, -2)\)
  • Let B \((x_2, y_2) = (5 , 1)\)
  • Let C \((x_3, y_3) = (3, -k)\)

\[\begin{align}\text{Area of a triangle} &= \frac{{\text{1}}}{{\text{2}}}\left\{ {{{\text{x}}_{\text{1}}}\left( {{{\text{y}}_{\text{2}}} - {{\text{y}}_{\text{3}}}} \right){\text{ + }}{{\text{x}}_{\text{2}}}\left( {{{\text{y}}_{\text{3}}} - {{\text{y}}_{\text{1}}}} \right){\text{ + }}{{\text{x}}_{\text{3}}}\left( {{{\text{y}}_{\text{1}}} - {{\text{y}}_{\text{2}}}} \right)} \right\}\,\\ &= \,0\ \text{for Collinear Points}\end{align}\]

By substituting the values of vertices, \(A, B, C\) in the Equation (1),

\[\begin{align}{\frac{{\text{1}}}{{\text{2}}}{\text{[7\{ 1}} - {\text{k\} + 5\{ k}} - {\text{(}} - {\text{2)\} + 3\{ (}} - {\text{2)}} - {\text{1\} ] = 0}}}\\{{\text{7}} - {\text{7k + 5k + 10}} - {\text{9 = 0}}}\\{ - {\text{2k + 8 = 0}}}\\{{\text{k = 4}}}\end{align}\]

Hence the given points are collinear for \(\begin{align}{\text{k = 4}}\end{align}\)

(ii) Given,

  • Let A \((x_1, y_1) = (8, 1)\)
  • Let B \((x_2, y_2) = (k , -4)\)
  • Let C \((x_3, y_3) = (2, -5)\)

\[\begin{align}\text{Area of a triangle}&=\frac{{{1}}}{{{2}}}\left\{ {{{{x}}_{{1}}}\left( {{{{y}}_{{2}}} - {{{y}}_{{3}}}} \right){{ + }}{{{x}}_{{2}}}\left( {{{{y}}_{{3}}} - {{{y}}_{{1}}}} \right){{ + }}{{{x}}_{{3}}}\left( {{{{y}}_{{1}}} - {{{y}}_{{2}}}} \right)} \right\}\,\\ &= \,0 \text{ for Collinear Points}\end{align}\]

By substituting the values of vertices, \(A, B, C\) in the Equation (1),

\[\begin{align} {\frac{{\text{1}}}{{\text{2}}}{\text{[8\{ }} - {4} - ( - {{5)\} + k\{ (}} - {{5)}} - {{(1)\} + 2\{ 1}} - {(} - {{4)\}]}}}&= 0\\ \text{8} - \text{6k + 10} &= 0 \left( {\text{By Transposing}} \right)\\ \text{6k} &= 18\\ \text{k}&= 3\end{align}\]

Hence the given points are collinear for \(\begin{align}{\text{k = 3}}\end{align}\)

  
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